organometallic chem

7/28/2019 Organometallic Chem.

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Organometallic Chemistry: (Ch. 15, pg 623-664, 705-709)HW: 3,4,5,7,8,38,52

Generally defined as compounds that possess ametal-carbon bond but there are exceptions to this definition.

Practical level: 25 Billion dollars were madefrom O-M processes in 1985.

Metal-carbonyls are the most typical O-M complexes.

Why is CO such a good ligand when it is such a poor lewis base?

Backbonding generates a metal-carbon bond.

How can we tell if an O-M compound will be reactive?

The 18-electron Rule helps us in 90% of the cases.The 18-electron rule is loosely based on the valence shell:

1-s, 3-p, and 5-d-orbitals = 9 orbitals total.

2e- each = 18 e-s.How to count the 18 electrons?

We will use the neutral atom method to count electrons.

1)The metal has all its electrons and is neutral.

2)The ligands donate a set number of electrons (Table 15.1).

3)The overall charge either adds e- or removes them.

The key feature of this counting method is knowinghow many electrons a particular ligand donates.

Lets start with the simple 2 e- ligands: carbonyl and phosphine.

Week 10-1


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Now the question is, are the COs terminal or bridging?

Two ways to see the difference:IR and structure.

IR:free CO 2150 cm -1

terminal

2000 cm-1

bridging 1800 cm -1.

How does the CO bridge the metals and still have back-bonding?

Week 10-2


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In these compounds, the Mn only shows 2000 peaks, while the Co shows 2000 and 1800 peaks. Structures confirm assignment.

Now lets consider ionic ligands, such as the halides.

The neutral atom counting method treats halogens, hydrogen and alkylligands as one electron neutral donors.

Think of these ligands as stong electrophiles so they only give upone electron to the metal.

Note that CO donates both electrons from its lone pair.

One can also count electrons with the Ox. State method.

Week 10-3


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However, note that when halogens bridge metals then they contribute 3electrons.

M-CL: + M = M-Cl-M

2-Rh = 18

4-CO = 8 32 total, 16e-/Rh2-Cl = 6Exception to the 18 e- rule

Hydrogen can be broken when reacting with certain complexes.

Co 2(CO) 8 + H 2 = 2 HCo(CO) 4 Co = 9e- Co = 9e-4CO = 8e- 4CO = 8e-Co- = 1e- H = 1e-

The hydrogen can either be thought of as H * or H -

Unsaturated organic ligandscan bind in two manners.

Note that one bonding method contributes 1electron while the other contributes 3 electrons.

Their names reflect this by being 1-allyl and 3-allyl.

This can also be seen forcyclopentadienyl (Cp).

Week 10-4


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The power of the 18 electron rule can be seen in the W(CO) 2(C 5H 5)2 complex.

If you count the electrons and assume that the Cps are5 coordinate then the total electrons are 20.

W = 6e-

2-Cp = 10e-2-CO = 4e-

The 18 electron rule would suggest thatone Cp only contributes 3 electrons and is bent.

This is seen in the xtal structure.

So, remember that the 18 electron rule is very helpfulbut it is ONLY a rule.

It helps us but does not have a formal explanation, similar to Lewis dot theory.

Metal Carbonyl Complexes: Why are they stable?

1) CO is not a strong Lewis basebut it forms strong bonds due to back-bonding.

2) Metals have to be in low oxidation state, often zero for this back-bonding.

3) Often obey the 18 electron rule.

How do we make these complexes?

Ni0 + 4CO (1 atm, 25 C) = Ni(CO) 4 + heat = Ni 0

However, iron requires more energy to make it react.Fe 0 + 5CO (200 atm, 200 C) = Fe(CO) 5

Some metals have to be reduced in the presence of CO.

CrCl 3 + Al + 6CO = AlCl 3 + Cr(CO) 6

Week 10-5


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Substitution reactions with carbonyl complexes:

Ni(CO) 4 + PR 3 = [Ni(CO) 3PR 3 + CO(g)]

Cr(CO) 6 + benzene = bisbenzenechromium(0) + 6 CO(g)

Reduction reactions:

Mn 2(CO) 10 + 2Na 0 = 2 Na[Mn(CO) 5]

Fe 3(CO) 12 + 6Na = 3 Na 2[Fe(CO) 4}2-

Note that the 18 electron rule is still obeyed.

Hydride Reactions:

The lone pair that is created by the above reductionreactions is very basic and is a good nucleophile.

Note that some organometallic lone pairs are so basic they candeprotonate organic solvents

Week 10-6


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Nitrosyl Chemistry:

Mixed carbonyl-nitrosyl complexes are known.

They form readily by CO replacement.

However, one can avoid the use of NO (toxic gas) by using NOCl.

NO + is isoelectronic with CO so it is not surprisingthat they are similar in their chemistry.

NO can bind through either the O or the N but it prefers to bindthrough the N to avoid a formal positive charge on the oxygen.

A difference between NO and CO is that NOcan bind to the metal bent and linear.

When it is bent, the N is in the sp 2 configuration.

In this form, it is considered a one electron donor.

Think of the radical sharing a bond with the metal.

The N is sp in the linear form.

In this form it is considered a three electron donor.

Think of both the lone pair and the radicalcontributing to the bond.

Week 10-7


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The first well characterized complexwith a bent NO was Vaskas complex.

So, how do we determine if the NO will be bent or linear?

The key point will be where will thelone pair live on the complex?

If the metal is electron poor, then there will benonbonding orbitals on the metal to accept thelone pair of the N and hence the NO will be linear (i.e. sp).

If the metal is electron-rich, then the metalorbitals will be filled and hence the nitrogenmust keep the lone pair (i.e. sp 2, bent config.).

Week 10-8


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This can be seen more clearly in the following comparison.

Dinitrogen Complexes

N2 is isoelectronic with CO and NO + but for years there were no N 2 complexes.

This was ascribed to the fact that N 2 is not polar and thus not a

good pi-acceptor.This is not the case for CO because its antibonding orbital isconcentrated on the carbon atom which favors overlap withthe metal.

Some of the first complexes with N 2 were with Ru.

Week 10-9


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Like oxygen, the nitrogen has a number of binding modes.

This is a critical process because it is the

key step of nitrogen fixation in nature.

N2 + 8H + + 8e- + 16 MgATP = 2 NH 3 + H 2 + 16 MgADP + 16 P i

Fertilizers: Natural source was bird guano from the islands off Peru.

Then a german, Haber, developed a process to make NH 3 and hence thebeginning of the green revolution.

Haber Process: N 2 + H 2 = 2 NH 3

Simple reaction but requires 1000 atm, 550 and Fe 0 /Fe 2O 3 to occur.

Metal Alkyls, Carbenes and Carbynes.

This means that one can form metal-carbonsingle, double and triple bonds.

A common reaction that can occur witha sigma-alkyl complex is a beta-elimination.

In this reaction, the metal pulls off a hydridefrom the bound alkyl, releasing a alkene.

Remember this last reaction because we will talk about the reverse soon.

Week 10-10


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One can also bind an alkene to a metal in a side-on fashion.

In this complex the pi orbital donates into thesigma bond of the metal and the metal d-orbitalsback-bond into the pi* orbital

Now lets use these simple complexes to help usunderstand an O-M catalytic reaction:

Alkene Hydrogenation

The first catalyst for this reaction was the Wilkinsons Catalyst.

Week 10-11


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The complex is a sq. planar, 16-electron d8 complex.

The two critical requirements for a good catalyst.

1)It needs to have an open coordination site.2)It cant be too stable (hence the 16-electrons).

As a ligand adds to the complex, the coordination numberincreases along with the oxidation state.

The concept is that as you add a ligand the metal has to contribute electronsto the new bonds being formed and hence is oxidized, oxidative addition.

It then follows that when the ligands leavethe metal is reduced to its original oxidation state.

[Pt(II)Cl 4]2- + Cl 2 = [Pt(IV)Cl 6]2-

Week 10-12


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How does the Wilkinson catalyst work?

First, a phosphine dissociates and a solvent molecules takes its place.This is because the phosphine is sterically large.

Then the hydrogen substrate binds, forming a cis-dihydride.

Then the second substrate, alkene, binds.

At this point, one of the hydrides migrates onto the alkene. This is the rate-determining step and forms a sigma-alkyl.

Solvent then fills the coordination environment (6C).

Finally, the last hydrogen goes on the substrate and you are leftwith the original catalyst.

Week 10-13


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This is better shown as a catalytic cycle.Week 10-14

organometallic chem

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